SOLUTION: Describe the collection of points in a plane so that the distance from each point to the point (5, 0) is five-fourths of its distance from the line x = 16/5

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Describe the collection of points in a plane so that the distance from each point to the point (5, 0) is five-fourths of its distance from the line x = 16/5       Log On


   

Question 1128793: Describe the collection of points in a plane so that the distance from each point to the point (5, 0) is five-fourths of its distance from the line x = 16/5
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The instruction "describe" is not specific; so let's just find the equation of the locus of points.

The distance of a point (x,y) from (5,0) is

sqrt%28%28x-5%29%5E2%2By%5E2%29

The distance of a point (x,y) from x = 16/5 is

abs%28x-16%2F5%29%29

The locus described is the set of points for which the distance from (5,0) is 5/4 the distance from x = 16/5:

sqrt%28%28x-5%29%5E2%2By%5E2%29+=+%285%2F4%29%28abs%28x-16%2F5%29%29%29
%28x-5%29%5E2%2By%5E2+=+%2825%2F16%29%28x%5E2-%2832%2F5%29x%2B256%2F25%29
x%5E2-10x%2B25%2By%5E2+=+%2825%2F16%29x%5E2-10x%2B16
%289%2F16%29x%5E2-y%5E2+=+9
x%5E2%2F16-y%5E2%2F9+=+1
x%5E2%2F4%5E2-y%5E2%2F3%5E2+=+1

Okay; so this is a recognizable locus.

The locus is a hyperbola with center at the origin; branches opening right and left; transverse axis (through the two vertices) length 2*4=8; conjugate axis (through the center perpendicular to the transverse axis) length 2*3 = 6.